131.

The values of the constants $\alpha$ and $\beta$ may be found from those of $V$ and $E$. For, on substituting the value of $w$ just found in [eqn:(178)] (178) and taking account of [eqn:(177)] (177) and [eqn:(182)] (182), we get $\frac{G}{m^3}=\alpha V{\iiint }_{-\infty }^{\infty }{e}^{-\beta ({\xi }^2+{\eta }^2+{\zeta }^2)}d\xi d\eta d\zeta =\alpha V{\left(\frac{\pi }{\beta }\right)}^{\frac{3}{2}},$

and on substituting $w$ in [eqn:(181)] (181) we get

$E=E_0+\frac{\alpha m^{4}NV}{2G}{\iiint }_{-\infty }^{\infty }({\xi }^2+{\eta }^2+{\zeta }^2)e^{-\beta ({\xi }^2+{\eta }^2+{\zeta }^2)}d\xi d\eta d\zeta ,$

or

$E=E_0+\frac{3\alpha m^{4}NV}{4G}\frac{1}{\beta }{\left(\frac{\pi }{\beta }\right)}^{\frac{3}{2}}.$

Solving for $\alpha$ and $\beta$ we have

$$\begin{gathered} \alpha \&=\frac{G}{V}{\left(\frac{3N}{4\pi m(E-E_0)}\right)}^{\frac{3}{2}}\mathrm{\backslash Label}[eqn](184)\text{upshape (184)} \\ \beta \&=\frac{3}{4}\frac{Nm}{E-E_0}.\mathrm{\backslash Label}[eqn](185)\text{upshape (185)} \end{gathered}$$

From this finally we find, as an expression for the entropy $S$ of the gas in the state of equilibrium with given values of $N$, $V$, and $E$, $S=kN\log \left\{\frac{V}{G}{\left(\frac{4\pi em(E-E_0)}{3N}\right)}^{\frac{3}{2}}\right\}.\mathrm{\backslash Label}[eqn](186)\text{upshape (186)}$